Setup
Let's focus on the ground state for concreteness. The wave function is
\begin{equation}
\psi_0(\vec{r}) = \frac{1}{\sqrt{\pi a_0^3} } e^{-|\vec{r}|/a_0}
\end{equation}
where $a_0$ is the Bohr radius.
Probability density function
Given the wavefunction, we can compute a three dimensional probability density function (pdf) $P(\vec{r})$
\begin{equation}
P_{3D}(\vec{r}) = |\psi_0(\vec{r})|^2 = \frac{1}{\pi a_0^3} e^{-2 |\vec{r}| / a_0}
\end{equation}
Note that the dimension of $P_{3D}$ is $({\rm length})^{-3}$.
Now what does this pdf tell us? Given a small volume of space, whose center is separated from the origin by vector $\vec{r}$, the probability that the electron is contained in that volume is $P_{3D}(\vec{r}) dV$, where $dV$ is the volume of that small volume element.
We can check that if we add up the probability associated with every volume element in space, we get $1$ (since we started from a properly normalized wavefunction).
\begin{eqnarray}
\int dV P_{3D}(\vec{r}) &=& \int_0^\infty dr r^2 \int_0^\pi d\theta \sin \theta\int_0^{2\pi} d\phi \frac{1}{\pi a_0^3} e^{-2r/a_0} \\
&=& 4\pi \times \frac{1}{\pi a_0^3} \int_0^\infty dr r^2 e^{-2r/a_0} \\
&=& 4\pi \times \frac{1}{\pi a_0^3} \times \frac{a_0^3}{4} \\
&=& 1
\end{eqnarray}
Radial probability density function
Now, sometimes, instead of wanting to know the probability of finding the electron in a small volume, we want to know what is the probability of finding the electron at some distance from the center. This is equivalent to asking about the probability of the electron being in a spherical shell of radius $r$.
You can derive this in various ways, but the upshot is that you derive a one-dimensional radial probability density $P_{1D}(r)$, which is related to the three-dimensional probability density $P_{3D}(\vec{r})$, via
\begin{equation}
P_{1D}(r) = 4\pi r^2 P_{3D}(\vec{r}) = \frac{4}{a_0^3} r^2 e^{-2 r/a_0}
\end{equation}
Note that the dimension of $P_{1D}$ is $({\rm length})^{-1}$ (which is different from the dimension of $P_{3D}$).
The interpretation of this function is that the probability to find an electron between distances $r$ and $r+dr$ from the origin is given by $P_{1D}(r)dr$.
We can check that the probability of finding the electron at some distance is $1$
\begin{eqnarray}
\int_0^\infty dr P_{1D}(r) &=& \frac{4}{a_0^3}\int_0^\infty dr r^2 e^{-2 r/a_0} \\
&=& \frac{4}{a_0^3} \times \frac{a_0^3}{4} \\
&=& 1
\end{eqnarray}
Let's also return to $P_{\rm 3D}(\vec{r})$ and see another way to interpret it. We said $P_{\rm 3D}(\vec{r}) dV$ was the probability to find the electron in an infinitesimal volume element with volume $dV$. Since $dV=r^2 dr d\Omega$, where $d\Omega=\sin\theta d\theta d\phi$ is an infinitesimal angular area element on the surface of a sphere, you could also express this probability as $P_{\rm 3D}(\vec{r}) r^2 dr d\Omega = P_{\rm 1D}(r) dr \frac{d\Omega}{4\pi}$. This is the same as the probability of finding the electron in a spherical shell of radius $r$, up to the normalization factor of $d\Omega/4\pi$ that converts the entire surface area of the shell to the infinitesimal cross section associated with the volume element with volume $dV$.
To confuse things futher, sometimes people may be lazy about normalizations, and, say, forget to put in a factor of $4\pi$ but still have in mind a radial pdf. Usually this is ok in practice because when one looks at plots of the probability density, one is comparing relative values and not absolute values. Hopefully, a good and careful resource will not do this.
